# Solving a rather unusual (diagonal) circuit

I have problems solving an electric circuit. I need to find the red drawn "I" (Current). However the way the electric circuit is drawn strikes me. I have very basic knowledge of physics and my skills are limited. The way I tried to attempt the problem is to look at the bottom left triangle and the top right triangle. I then used the Mesh-Current-Method for the bottom left triangle which I defined as: $m_1 = -U_q_1 + U_1 + U_3 + U_q_3$ ($U_1$ is the Voltage at $R_1$ and $U_3$ the Voltage at $R_3$). This term must be equal 0 due to Kirchoff law. Knowing $U_q_1$ and $U_q_3$ (both 5V) we are left with $m_1 = U_1 + U_3$. As far as I know $U_1$ and $U_3$ cannot be negative, so I assumed that both are 0V (due to the addition). Repeating this for the upper triangle returns me $U_2 = 0$. With Ohm's law $I$ must be 0. However my own intuition is that I did something wrong. Any ideas?

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Superposition is a great tool to solve problems as another answer shows. However, in this case, there is a simpler method.

Note that the leftmost branch and the diagonal branch are not only identical, they are in parallel. Thus the voltage across each branch is identical and, since each branch is identical, the current through each branch has the same value.

The implication is that both branches can be replaced with an equivalent circuit that is just a 5V source in series with a 1/2 ohm resistor.

Having done this, finding I is trivial as you're left with a simple series circuit to solve: I = - 10V / 2.5 ohms = -4A

UPDATE: According to the OP's comment below, the answer is +4A. This can only be true if the arrows associated with the voltage sources in the diagram are drawn from the positive terminal to the negative terminal.

The voltage source symbols used in the OP's diagram, however, are ambiguous.

Typically, a DC voltage source has a "plus" sign to denote the positive terminal with an optional arrow as so:

Note that the arrow is drawn from the negative to the positive. If the OP's diagram follows this convention, the correct answer is -4A, not +4A.

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Today I got the solution for this circuit and the missing I is positive 4A not negative 4A. I have no clue in how far that is correct but I just wanted to share with you. –  optional Nov 23 '12 at 18:41
@optional, I've updated the answer to address your comment. –  Alfred Centauri Nov 23 '12 at 21:51

As a general rule, clearly define all voltage (including polarities). This can be done by labeling either each component of the schematic with the a polarity (+ or $\rightarrow$). Alternatively, you can define a 'reference' for each node in the schematic (eg: by a letter), as shown below.

We will indicate the voltage of $A$ with respect to $B$ as $V_{AB}$.

For example, $U_{q1}=V_{FD}=-V_{DF}$

Now, applying Kirchhoff's Voltage Law (KVL) around loop A-E-F-D-A, we get:

$V_{AE}+V_{EF}+V_{FD}+V_{DA}=0$

ie:

$I_{3}R_{3}-U_{q3}+U_{q1}-I_{1}R_{1}=0$

Substituting known value, we get

$I_{1}=I_{3}$

Applying Kirchhoff's Current Law (KCL) at node A,

$I_{1}+I_{3}-I_{4}=0$

Which gives,

$I_{4}=2I_{1}=2I_{3}$, or $I_{3}=I_{4}/2$

Finally, applying KVL around loop A-B-C-F-E-A, we get:

$V_{AB}+V_{BC}+V_{CF}+V_{FE}+V_{EA}=0$

ie

$-I_{4}R_{2}+U_{q2}-I_{4}R_{4}+U_{q3}-I_{3}R_{3}=0$

Substituting known values,

$-I_{4}+5-I_{4}+5-I_{4}/2=0$

ie

$I_{4}=4 A$

so $I = -I_{4} = -4 A$

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I've got a question concerning the calculation of the second loop. If one applys KVL around loop A-B-C-F-E-A isn't $U_q3$ pointing in the opposite direction and $I_4$ towards A? As I stated in another question I am not very skilled in physics but having given the law shouldn't $V_AB + V_BC + V_CF + V_FE + V_EA = 0$ actually be: $I_4R_2 + U_q2 + I_4R_4 - U_q3 - I_3R_3$ which would result in $I_4 = 0$. However my intuition tells me that as you state above $V_AB$, which indicates the Voltage of A with respect to B should be considered seperate thus $I_4$ must be negative in that relation. @theo –  optional Nov 21 '12 at 9:44
The arrow on $U_{q3}$ is indicating that node F is 5V higher than node E. That is, the potential of F with respect to E is 5V, or $V_{FE}=5$. In the case of the current $I_{4}$ through resistor $R_{2}$, the voltage at A with respect to B is negative of the voltage of B with respect to A, where $V_{BA}=I_{4}R_{2}$. That is, if $I_{4}$ indicates positive current flow in the direction shown, the potential at B will be higher than the potential at A, so $V_{BA}$ will be positive. –  theo Nov 21 '12 at 12:21

When you have circuits with multiple sources you should use the principle of superposition to solve a linear problem such as this one.

If you turn off $U_{q1}$ and $U_{q2}$ but not $U_{q3}$, you can easily solve the value for $I$, let's call it $I_3$.

Then you do this same step for the two other sources:

1. you turn off $U_{q1}$ and $U_{q3}$ and then solve $I$, called $I_2$ ;

2. you turn off $U_{q2}$ and $U_{q3}$ and then solve $I$, called $I_1$.

The solution for $I$ when the three sources are turned on then is $$I = I_1 + I_2 + I_3$$

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I am not quite sure how to attempt the problem when turning off $U_q1$ and $U_q2$ for example. Is the Voltage at every resistor equal to $U_q3$? My problem is that I have 2 unknown and 1 equation. –  optional Nov 21 '12 at 9:59